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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-programming-design-and-analysis-of-algorithms-resit/2/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms (resit) / 2 / b / Solution by

Codex  0 2026-10-31

Use merge sort, then project the second component of each pair. Under the ordinary Haskell Ord class instance for pairs, “ascending” means lexicographic order, requiring both component types to be ordered:

sort :: (Ord a, Ord b) => [(a,b)] -> [b]
sort = map snd . mergeSort

mergeSort :: Ord a => [a] -> [a]
mergeSort [] = []
mergeSort [x] = [x]
mergeSort xs = merge (mergeSort left) (mergeSort right)
  where (left,right) = splitAt (length xs `div` 2) xs

merge :: Ord a => [a] -> [a] -> [a]
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys)
  | x <= y    = x : merge xs (y:ys)
  | otherwise = y : merge (x:xs) ys

If the intended convention is to sort only by the first component as a sort key, replace x <= y by fst x <= fst y and specialize the merge routines to pairs. The weaker signature is then sort :: Ord a => [(a,b)] -> [b], and equal keys keep their original order. These conventions differ when first components tie; the question does not explicitly specify that tie rule.

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